It is a graph that shows the relationship between the quantity demanded of a commodity at different prices over a given period of time. It is observed that the demand curve slopes downward from left to right. It shows it has a negative slope which implies that consumers purchase more of commodity at lower prices than at higher prices.
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<span>You are given an annual dividend of $2.10 for the fifteen years that you plan on holding it. Also, after 15 years, you are given to sell the stock for $32.25. You are asked to find the present value of a share for this company if you want a 10% return. You have to mind that the future stock for 15 years is $32.25. You are not only going to mind the present value of the annuity at $2.10 but also the $32.25.
With the interest of r = 10% and number of years of n = 15, we get
PVIFA = 7.6061.
For annuity we have,
$2.10 * 7.60608 = $15.973
For $32.35 with r = 10% and n = 15
PVIF = 0.239392
Thus for the present value of selling price,
$32.25 * 0.239392 = $7.720
Thus the present value of the share
P = $15.973 + $7.720
P = $23.693
</span>
Answer:
The answer is below.
Explanation:
The z score is a used in statistics to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
![z=\frac{x-\mu}{\sigma}\\\\where\ x=raw\ score, \mu=mean,\sigma=standard\ deviation\\\\For\ a\ sample\ size(n):\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5C%5C%5C%5Cwhere%5C%20x%3Draw%5C%20score%2C%20%5Cmu%3Dmean%2C%5Csigma%3Dstandard%5C%20deviation%5C%5C%5C%5CFor%5C%20a%5C%20sample%5C%20size%28n%29%3A%5C%5C%5C%5Cz%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%20%7D)
a) Given that n = 100, μ = 2000, σ = 18
For x < 1995 millimeters:
![z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{1995-2000}{18/\sqrt{100} } =-2.78](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%20%7D%3D%5Cfrac%7B1995-2000%7D%7B18%2F%5Csqrt%7B100%7D%20%7D%20%20%3D-2.78)
From the normal distribution table, P(x < 1995) = P(z < -2.78) = 0.0027
b) P(z > z*) = 10% = 0.1
P(z < z*) = 1 - 0.1 = 0.9
z* = 1.28
![z*=\frac{x-\mu}{\sigma/\sqrt{n} }\\\\1.28=\frac{x-2000}{18/\sqrt{100} }\\\\x-2000 =-2.304\\\\x=2002.3\ ml\\\\](https://tex.z-dn.net/?f=z%2A%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%20%7D%5C%5C%5C%5C1.28%3D%5Cfrac%7Bx-2000%7D%7B18%2F%5Csqrt%7B100%7D%20%7D%5C%5C%5C%5Cx-2000%20%20%3D-2.304%5C%5C%5C%5Cx%3D2002.3%5C%20ml%5C%5C%5C%5C)
From the normal distribution table, P(z < z
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